\(a,4x^2-6x=2x\left(2x-3\right)\\ b,9x^4y^3+3x^2y^4=3x^2y^3\left(2x^2+y\right)\\ c,x^3-2x^2+5x=x\left(x^2-2x+5\right)\\ d,3x\left(x-1\right)+5\left(x-1\right)=\left(3x+5\right)\left(x-1\right)\\ e,2x^2\left(x+1\right)+4\left(x+1\right)=\left(x+1\right)\left(2x^2+4\right)=2\left(x+1\right)\left(x^2+2\right)\\ f,2x^2y-4xy^2+6xy=2xy\left(x-y+3\right)\\ g,4x^3+4x^2+4x=4x\left(x^2+x+1\right)\\ h,x^3+x^2-3x-27=x^3-3x^2+4x^2-12x+9x-27=x^2\left(x-3\right)+4x\left(x-3\right)+9\left(x-3\right)=\left(x^2+4x+9\right)\left(x-3\right)\\ i,4x^2-12x+9=\left(2x-3\right)^2\\ k,8x^3-27=\left(2x\right)^3-3^3=\left(2x-3\right)\left(4x^2+6x+9\right)\\ l,x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)

Tick nha 😘

\(a,\left(x+1\right)^3+x\left(x-2\right)^2+x-1=0\\ \Leftrightarrow x^3+3x^2+3x+1+x\left(x^2-4x+4\right)+x-1=0\\ \Leftrightarrow x^3+3x^2+3x+1+x^3-4x^2+4x+x-1=0\\ \Leftrightarrow2x^3-x^2+8x=0\\ \Leftrightarrow x\left(2x^2-x+8\right)=0\\ \Leftrightarrow x=0\)

Do \(2x^2-x+8=2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}-\dfrac{1}{16}+4\right)=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{63}{8}>0\)

1.have lived

2.lost

3.sank

4.have owned

5.has been

6.cried

7.have bought

8.ate

9.has fallen

10.have drunk

Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)

Khi đó phương trình trở thành: 

\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)

Tick plz

\(1,\) Khi \(x=9\) thì \(A=\dfrac{x+\sqrt{x}+4}{\sqrt{x}-2}=\dfrac{9+\sqrt{9}+4}{\sqrt{9}-2}=16\)

\(2,\) Ta có \(B=\dfrac{3x-4}{x-2\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}}+\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\left(x>0;x\ne4\right)\)

 \(B=\dfrac{3x-4-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ B=\dfrac{3x-4-x+4-x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)

kkk khá đau đầu 😄

Đề bạn có mấy chỗ thiếu mk bổ sung nha

\(a,2^3+4^2+6x=8+16+6x=6x+24=x\left(x+4\right)\\ b,x^2-4=\left(x-2\right)\left(x+2\right)\\ c,x^2-10x+25=\left(x-5\right)^2\\ d,x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\\ e,x^2+xy-3x-3y=x\left(x+y\right)-3\left(x+y\right)=\left(x-3\right)\left(x+y\right)\\ g,x^2-y^2-4x+4=\left(x-2\right)^2-y^2=\left(x-y-2\right)\left(x+y-2\right)\)

Tick plzz

P:   Aa    ×   Aa
      G  : A; a      A; a                        
     F1 :  1AA : 2Aa : 1aa 

Tick plz

\(A=\sqrt{m^2+2m+1}+\sqrt{m^2-2m+1}\\ A=\sqrt{\left(m+1\right)^2}+\sqrt{\left(m-1\right)^2}\\ A=\left|m+1\right|+\left|m-1\right|=\left|m+1\right|+\left|1-m\right|\ge\left|m+1+1-m\right|=\left|2\right|=2\)

Dấu "=" \(\Leftrightarrow m+1=1-m\Leftrightarrow m=0\)

\(B=\sqrt{4a^2-4a+1}+\sqrt{4a^2-12a+9}\\ B=\sqrt{\left(2a-1\right)^2}+\sqrt{\left(2a-3\right)^2}\\ A=\left|2a-1\right|+\left|2a-3\right|=\left|2a-1\right|+\left|3-2a\right|\ge\left|2a-1+3-2a\right|=\left|2\right|=2\)

Dấu "=" \(\Leftrightarrow2a-1=3-2a\Leftrightarrow a=1\)

Tick plz

Pt hoành độ giao điểm:

Vậy